Contents

Data format: Contingency tables

We’ll use housetasks data sets from STHDA: https://www.sthda.com/sthda/RDoc/data/housetasks.txt.

# Import the data
file_path <- "https://www.sthda.com/sthda/RDoc/data/housetasks.txt"
housetasks <- read.delim(file_path, row.names = 1)
# head(housetasks)

An image of the data is displayed below:

Data format correspondence analysis

Graphical display of contengency tables

Contingency table can be visualized using the function balloonplot() [in gplots package]. This function draws a graphical matrix where each cell contains a dot whose size reflects the relative magnitude of the corresponding component.

To execute the R code below, you should install the package gplots: install.packages(“gplots”).

library("gplots")
# 1. convert the data as a table
dt <- as.table(as.matrix(housetasks))
# 2. Graph
balloonplot(t(dt), main ="housetasks", xlab ="", ylab="",
            label = FALSE, show.margins = FALSE)

Chi-Square Test of Independence in R

Note that, row and column sums are printed by default in the bottom and right margins, respectively. These values can be hidden using the argument show.margins = FALSE.

It’s also possible to visualize a contingency table as a mosaic plot. This is done using the function mosaicplot() from the built-in R package garphics:

library("graphics")
mosaicplot(dt, shade = TRUE, las=2,
           main = "housetasks")

Chi-Square Test of Independence in R

• The argument shade is used to color the graph
• The argument las = 2 produces vertical labels

Note that the surface of an element of the mosaic reflects the relative magnitude of its value.

• Blue color indicates that the observed value is higher than the expected value if the data were random
• Red color specifies that the observed value is lower than the expected value if the data were random

From this mosaic plot, it can be seen that the housetasks Laundry, Main_meal, Dinner and breakfeast (blue color) are mainly done by the wife in our example.

There is another package named vcd, which can be used to make a mosaic plot (function mosaic()) or an association plot (function assoc()).

# install.packages("vcd")
library("vcd")
# plot just a subset of the table
assoc(head(dt, 5), shade = TRUE, las=3)

Chi-Square Test of Independence in R

Chi-square test basics

Chi-square test examines whether rows and columns of a contingency table are statistically significantly associated.

• Null hypothesis (H0): the row and the column variables of the contingency table are independent.
• Alternative hypothesis (H1): row and column variables are dependent

For each cell of the table, we have to calculate the expected value under null hypothesis.

For a given cell, the expected value is calculated as follow:

The Chi-square statistic is calculated as follow:

This calculated Chi-square statistic is compared to the critical value (obtained from statistical tables) with \(df = (r - 1)(c - 1)\) degrees of freedom and p = 0.05.

• r is the number of rows in the contingency table
• c is the number of column in the contingency table

If the calculated Chi-square statistic is greater than the critical value, then we must conclude that the row and the column variables are not independent of each other. This implies that they are significantly associated.

Note that, Chi-square test should only be applied when the expected frequency of any cell is at least 5.

Compute chi-square test in R

Chi-square statistic can be easily computed using the function chisq.test() as follow:

chisq <- chisq.test(housetasks)
chisq

    Pearson's Chi-squared test

data:  housetasks
X-squared = 1944.5, df = 36, p-value < 2.2e-16

In our example, the row and the column variables are statistically significantly associated (p-value = 0).

The observed and the expected counts can be extracted from the result of the test as follow:

# Observed counts
chisq$observed

           Wife Alternating Husband Jointly
Laundry     156          14       2       4
Main_meal   124          20       5       4
Dinner       77          11       7      13
Breakfeast   82          36      15       7
Tidying      53          11       1      57
Dishes       32          24       4      53
Shopping     33          23       9      55
Official     12          46      23      15
Driving      10          51      75       3
Finances     13          13      21      66
Insurance     8           1      53      77
Repairs       0           3     160       2
Holidays      0           1       6     153

# Expected counts
round(chisq$expected,2)

            Wife Alternating Husband Jointly
Laundry    60.55       25.63   38.45   51.37
Main_meal  52.64       22.28   33.42   44.65
Dinner     37.16       15.73   23.59   31.52
Breakfeast 48.17       20.39   30.58   40.86
Tidying    41.97       17.77   26.65   35.61
Dishes     38.88       16.46   24.69   32.98
Shopping   41.28       17.48   26.22   35.02
Official   33.03       13.98   20.97   28.02
Driving    47.82       20.24   30.37   40.57
Finances   38.88       16.46   24.69   32.98
Insurance  47.82       20.24   30.37   40.57
Repairs    56.77       24.03   36.05   48.16
Holidays   55.05       23.30   34.95   46.70

Nature of the dependence between the row and the column variables

As mentioned above the total Chi-square statistic is 1944.456196.

If you want to know the most contributing cells to the total Chi-square score, you just have to calculate the Chi-square statistic for each cell:

\[ r = \frac{o - e}{\sqrt{e}} \]

The above formula returns the so-called Pearson residuals (r) for each cell (or standardized residuals)

Cells with the highest absolute standardized residuals contribute the most to the total Chi-square score.

Pearson residuals can be easily extracted from the output of the function chisq.test():

round(chisq$residuals, 3)

             Wife Alternating Husband Jointly
Laundry    12.266      -2.298  -5.878  -6.609
Main_meal   9.836      -0.484  -4.917  -6.084
Dinner      6.537      -1.192  -3.416  -3.299
Breakfeast  4.875       3.457  -2.818  -5.297
Tidying     1.702      -1.606  -4.969   3.585
Dishes     -1.103       1.859  -4.163   3.486
Shopping   -1.289       1.321  -3.362   3.376
Official   -3.659       8.563   0.443  -2.459
Driving    -5.469       6.836   8.100  -5.898
Finances   -4.150      -0.852  -0.742   5.750
Insurance  -5.758      -4.277   4.107   5.720
Repairs    -7.534      -4.290  20.646  -6.651
Holidays   -7.419      -4.620  -4.897  15.556

Let’s visualize Pearson residuals using the package corrplot:

library(corrplot)
corrplot(chisq$residuals, is.cor = FALSE)

Chi-Square Test of Independence in R

For a given cell, the size of the circle is proportional to the amount of the cell contribution.

The sign of the standardized residuals is also very important to interpret the association between rows and columns as explained in the block below.

The contribution (in %) of a given cell to the total Chi-square score is calculated as follow:

• r is the residual of the cell

# Contibution in percentage (%)
contrib <- 100*chisq$residuals^2/chisq$statistic
round(contrib, 3)

            Wife Alternating Husband Jointly
Laundry    7.738       0.272   1.777   2.246
Main_meal  4.976       0.012   1.243   1.903
Dinner     2.197       0.073   0.600   0.560
Breakfeast 1.222       0.615   0.408   1.443
Tidying    0.149       0.133   1.270   0.661
Dishes     0.063       0.178   0.891   0.625
Shopping   0.085       0.090   0.581   0.586
Official   0.688       3.771   0.010   0.311
Driving    1.538       2.403   3.374   1.789
Finances   0.886       0.037   0.028   1.700
Insurance  1.705       0.941   0.868   1.683
Repairs    2.919       0.947  21.921   2.275
Holidays   2.831       1.098   1.233  12.445

# Visualize the contribution
corrplot(contrib, is.cor = FALSE)

Chi-Square Test of Independence in R

The relative contribution of each cell to the total Chi-square score give some indication of the nature of the dependency between rows and columns of the contingency table.

It can be seen that:

1. The column “Wife” is strongly associated with Laundry, Main_meal, Dinner
2. The column “Husband” is strongly associated with the row Repairs
3. The column jointly is frequently associated with the row Holidays

Access to the values returned by chisq.test() function

The result of chisq.test() function is a list containing the following components:

The format of the R code to use for getting these values is as follow:

# printing the p-value
chisq$p.value
# printing the mean
chisq$estimate

See also

• One Proportion Z-Test in R: Compare an Observed Proportion to an Expected One
• Two Proportions Z-Test in R: Compare Two Observed Proportions
• Chi-Square Goodness of Fit Test in R: Compare Multiple Observed Proportions to Expected Probabilities

Infos

This analysis has been performed using R software (ver. 3.2.4).

What is chi-square goodness of fit test?

Example data and questions

For example, we collected wild tulips and found that 81 were red, 50 were yellow and 27 were white.

1. Question 1:

Are these colors equally common?

If these colors were equally distributed, the expected proportion would be 1/3 for each of the color.

2. Question 2:

Suppose that, in the region where you collected the data, the ratio of red, yellow and white tulip is 3:2:1 (3+2+1 = 6). This means that the expected proportion is:

• 3/6 (= 1/2) for red
• 2/6 ( = 1/3) for yellow
• 1/6 for white

We want to know, if there is any significant difference between the observed proportions and the expected proportions.

Statistical hypotheses

• Null hypothesis (\(H_0\)): There is no significant difference between the observed and the expected value.
• Alternative hypothesis (\(H_a\)): There is a significant difference between the observed and the expected value.

R function: chisq.test()

The R function chisq.test() can be used as follow:

chisq.test(x, p)

tulip <- c(81, 50, 27)
res <- chisq.test(tulip, p = c(1/3, 1/3, 1/3))
res

    Chi-squared test for given probabilities

data:  tulip
X-squared = 27.886, df = 2, p-value = 8.803e-07

# Access to the expected values
res$expected

[1] 52.66667 52.66667 52.66667

tulip <- c(81, 50, 27)
res <- chisq.test(tulip, p = c(1/2, 1/3, 1/6))
res

    Chi-squared test for given probabilities

data:  tulip
X-squared = 0.20253, df = 2, p-value = 0.9037

# printing the p-value
res$p.value

[1] 0.9036928

# printing the mean
res$estimate

NULL

See also

• One Proportion Z-Test in R: Compare an Observed Proportion to an Expected One
• Two Proportions Z-Test in R: Compare Two Observed Proportions
• Chi-Square Test of Independence in R: Evaluate The Association Between Two Categorical Variables

Infos

This analysis has been performed using R software (ver. 3.2.4).

1 How this chapter is organized?

2 One-proportion z-Test

Read more: —> One-Proportion Z-Test in R.

3 Two-proportions z-Test

Read more: —> Two Proportions Z-Test.

4 Chi-square goodness of fit test in R

Read more: —> Chi-square goodness of fit test in R.

5 Chi-Square test of independence in R

Read more: —> Chi-Square Test of Independence in R.

6 See also

• R Basics
• Import and Export Data using R
• Preparing and Reshaping Data in R for Easier Analyses
• Data Manipulation in R
• Data visualization
• Descriptive Statistics and Graphics
• Correlation Analyses in R
• Comparing Means in R
• Comparing Variances in R

7 Infos

This analysis has been performed using R statistical software (ver. 3.2.4).

What is two-proportions z-test?

For example, we have two groups of individuals:

• Group A with lung cancer: n = 500
• Group B, healthy individuals: n = 500

The number of smokers in each group is as follow:

• Group A with lung cancer: n = 500, 490 smokers, \(p_A = 490/500 = 98%\)
• Group B, healthy individuals: n = 500, 400 smokers, \(p_B = 400/500 = 80%\)

In this setting:

• The overall proportion of smokers is \(p = frac{(490 + 400)}{500 + 500} = 89%\)
• The overall proportion of non-smokers is \(q = 1-p = 11%\)

We want to know, whether the proportions of smokers are the same in the two groups of individuals?

Research questions and statistical hypotheses

Typical research questions are:

In statistics, we can define the corresponding null hypothesis (\(H_0\)) as follow:

1. \(H_0: p_A = p_B\)
2. \(H_0: p_A \leq p_B\)
3. \(H_0: p_A \geq p_B\)

The corresponding alternative hypotheses (\(H_a\)) are as follow:

1. \(H_a: p_A \ne p_B\) (different)
2. \(H_a: p_A > p_B\) (greater)
3. \(H_a: p_A < p_B\) (less)

Formula of the test statistic

Compute two-proportions z-test in R

prop.test(x, n, p = NULL, alternative = "two.sided",
          correct = TRUE)

res <- prop.test(x = c(490, 400), n = c(500, 500))

# Printing the results
res 

    2-sample test for equality of proportions with continuity correction

data:  c(490, 400) out of c(500, 500)
X-squared = 80.909, df = 1, p-value < 2.2e-16
alternative hypothesis: two.sided
95 percent confidence interval:
 0.1408536 0.2191464
sample estimates:
prop 1 prop 2 
  0.98   0.80 

prop.test(x = c(490, 400), n = c(500, 500),
           alternative = "less")

prop.test(x = c(490, 400), n = c(500, 500),
              alternative = "greater")

# printing the p-value
res$p.value

[1] 2.363439e-19

# printing the mean
res$estimate

prop 1 prop 2 
  0.98   0.80 

# printing the confidence interval
res$conf.int

[1] 0.1408536 0.2191464
attr(,"conf.level")
[1] 0.95

See also

• One Proportion Z-Test in R: Compare an Observed Proportion to an Expected One
• Chi-Square Goodness of Fit Test in R: Compare Multiple Observed Proportions to Expected Probabilities
• Chi-Square Test of Independence in R: Evaluate The Association Between Two Categorical Variables

Infos

This analysis has been performed using R software (ver. 3.2.4).

What is one-proportion Z-test?

For example, we have a population of mice containing half male and have female (p = 0.5 = 50%). Some of these mice (n = 160) have developed a spontaneous cancer, including 95 male and 65 female.

We want to know, whether the cancer affects more male than female?

In this setting:

• the number of successes (male with cancer) is 95
• The observed proportion (\(p_o\)) of male is 95/160
• The observed proportion (\(q\)) of female is \(1 - p_o\)
• The expected proportion (\(p_e\)) of male is 0.5 (50%)
• The number of observations (\(n\)) is 160

Research questions and statistical hypotheses

Typical research questions are:

In statistics, we can define the corresponding null hypothesis (\(H_0\)) as follow:

1. \(H_0: p_o = p_e\)
2. \(H_0: p_o \leq p_e\)
3. \(H_0: p_o \geq p_e\)

The corresponding alternative hypotheses (\(H_a\)) are as follow:

1. \(H_a: p_o \ne p_e\) (different)
2. \(H_a: p_o > p_e\) (greater)
3. \(H_a: p_o < p_e\) (less)

Formula of the test statistic

The test statistic (also known as z-test) can be calculated as follow:

\[ z = \frac{p_o-p_e}{\sqrt{p_oq/n}} \]

where,

• \(p_o\) is the observed proportion
• \(q = 1-p_o\)
• \(p_e\) is the expected proportion
• \(n\) is the sample size

The confidence interval of \(p_o\) at 95% is defined as follow:

\[ p_o \pm 1.96\sqrt{\frac{p_oq}{n}} \]

Note that, the formula of z-statistic is valid only when sample size (\(n\)) is large enough. \(np_o\) and \(nq\) should be \(\geq\) 5. For example, if \(p_o = 0.1\), then \(n\) should be at least 50.

Compute one proportion z-test in R

binom.test(x, n, p = 0.5, alternative = "two.sided")

prop.test(x, n, p = NULL, alternative = "two.sided",
          correct = TRUE)

res <- prop.test(x = 95, n = 160, p = 0.5, 
                 correct = FALSE)

# Printing the results
res 

    1-sample proportions test without continuity correction

data:  95 out of 160, null probability 0.5
X-squared = 5.625, df = 1, p-value = 0.01771
alternative hypothesis: true p is not equal to 0.5
95 percent confidence interval:
 0.5163169 0.6667870
sample estimates:
      p 
0.59375 

prop.test(x = 95, n = 160, p = 0.5, correct = FALSE,
           alternative = "less")

prop.test(x = 95, n = 160, p = 0.5, correct = FALSE,
              alternative = "greater")

# printing the p-value
res$p.value

[1] 0.01770607

# printing the mean
res$estimate

      p 
0.59375 

# printing the confidence interval
res$conf.int

[1] 0.5163169 0.6667870
attr(,"conf.level")
[1] 0.95

See also

• Two Proportions Z-test in R: Compare Two Observed Proportions
• Chi-Square Goodness of Fit Test in R: Compare Multiple Observed Proportions to Expected Probabilities
• Chi-Square Test of Independence in R: Evaluate The Association Between Two Categorical Variables

Infos

This analysis has been performed using R software (ver. 3.2.4).